Question

Hotels, like airline, often overbook, counting on the fact that some people with reservations will cancel at the last minute. A certain hotel chain finds 20% of the reservations will not be used. Let the Binomial random variable X be the number of reservations canceled. If we randomly selected 15 reservations, what is the probability that more than 8 but less than 12 reservations will be used?

Answer 1

Answer:

Probability that more than 8 but less than 12 reservations will be used is 0.334 .

Step-by-step explanation:

We are given that a certain hotel chain finds 20% of the reservations will not be used which means that 20% of the reservations will be used.

Let X = number of reservations canceled

X ~ $Binom(n=15,p = 0.80)$

The Probability distribution of Binomial distribution is given by;

$P(X=r) = \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

where, n = number of trials(samples) taken = 15

            r = number of success

            p = probability of success i.e. success of reservations will be used

So, probability that more than 8 but less than 12 reservations will be used is given by = P(X = 9) + P(X = 10) + P(X = 11)

     = $\binom{15}{9}0.8^{9}(1-0.8)^{15-9} + \binom{15}{10}0.8^{10}(1-0.8)^{15-10} + \binom{15}{11}0.8^{11}(1-0.8)^{15-11}$

     = $5005*0.8^{9} *0.2^{6} + 3003*0.8^{10} *0.2^{5} + 1365*0.8^{11} *0.2^{4}$

     = 0.043 + 0.103 + 0.188 = 0.334

Therefore, probability that more than 8 but less than 12 reservations will be used is 0.334 .

Answer 2

Answer: The required probability is 0.334.

Step-by-step explanation:

Since we have given that

N = 15 reservations

probability of reservations will not be used = 0.2

probability of reservations will be used = 1-0.2=0.8

Let X be the Binomial variable,that the number of reservations canceled.

We need to find the probability that more than 8 but less than 12 reservations.

So, it becomes,

$P(8$

Hence, the required probability is 0.334.